25 April 2 Page of 6 (Document Released: 4:3, 4/24) Theoretical Question 2: Strong Resistive Electromagnets SOLUTION Part A. Magnetic Fields on the Axis of the Coil (a) At the point xx on the axis, the magnetic field due to the current II passing through the turns located in the interval (ss, ss + ) is (see Fig. A) ddbb = ( μμ 4ππ ) II(ππππ) (DD/2) 2 + (ss xx) 2 (DD/2) (DD/2) 2 + (ss xx) 2 aa xx (a-) which, when summed over all turns of the coil, leads to the total magnetic field BB (xx) = BB(xx)xx with BB(xx) = μμ II 2aa (DD 2 )2 = μμ II 2aa ( DD 2 )2 l/2 l/2 [(DD/2) 2 + (ss xx) 2 ] 3/2 l/2 xx l/2 xx [(DD/2) 2 + ss 2 ] 3/2 = μμ II 2aa (l/2) xx (DD/2) 2 + [(l/2) xx] + (l/2) + xx 2 (DD/2) 2 + [(l/2) + xx] 2 no. of turns in is /aa (a-2)* Figure A 2 DD 2 DD OO xx ss ss + xx II (b) From Eq. (a-2), the magnetic field at O wth xx = is BB() = μμ II 2aa 2(l/2) (DD/2) 2 + (l/2) = μμ II 2 aa + (DD/l) 2 (b-) If BB() is. T, then the current II must be equal to II = BB() aa μμ + (DD/l) 2 =.7794 4 A.8 4 A (b-2)* ---------------------------------------------------------------------------------------------------------------- *An equation marked with an asterisk gives key answers to the problem.
25 April 2 Page 2 of 6 (Document Released: 4:3, 4/24) Part B. The Upper Limit of Current (c) For an infinitely long coil with l and bb DD, the magnetic field BB acting on the current is the average of the fields inside and outside of the coil. The field outside is zero and the field inside is the same as that at O, i.e. BB() in Eq. (b-) with l. Thus we have BB = BB xx = 2 ( + μμ II aa )xx = μμ II 2aa xx, (c-) and the outward normal force on the wire segment of length ss is FF n = IIBB ss = II ss μμ II FF n or 2aa ss = μμ 2aa II2. (c-2)* As can be seen from Fig. A2, the resultant of the pair of tension forces at the ends of the segment ss is given by 2FF t sin ( θθ 2 ) FF t θθ = FF t 2 ss DD. (c-3) This must be in equilibrium with the normal force FF n so that, by using Eq. (c-2), we have FF n = FF t 2 ss DD or FF t = DD 2 FF n ss = μμ 4aa II2 DD. (c-4)* FF n ss Figure A2 FF t bb θθ xx 2 DD II FF t (d) At breaking, the tensile stress of the wire is, from Eq. (c-4), FF t aaaa = μμ 4aa 2 bb II b 2 DD = σσ b = 4.55 8 Pa, (d-) and the tensile strain of the wire is ππ(dd DD) (DD DD) = = 6 % or DD =.6 DD. (d-2) ππππ DD From the last two equations, the current II b at which the turn will break is II b = 2aa bbbb b μμ DD = 2aa bbbb b μμ (.6 DD) =.737 4 A.7 4 A, and the magnitude of the magnetic field at the center O, i.e. Eq. (b-) with l, is (d-3) BB b = μμ II b aa = 2 μμ bbbb b DD =.94 T =. T, (d-4)*
25 April 2 Page 3 of 6 (Document Released: 4:3, 4/24) Part C. Rate of Temperature Rise (e) When the current II is. ka, the current density JJ is given by JJ = II aaaa =. 4 (2. 3 )(5. 3 ) =. 9 A/m 2. (e-) The power density is given by ρρ e JJ 2 = ρρ e II 2 aaaa =.72 W/m 3.7 W/m 3. (e-2)* (ALTERNATIVE) The volume ττ and resistance RR (appearing also in Problem (h)) of the current-carrying wire for a coil of length l are given by 2 2 DD + bb ττ = ππ 2 DD bb 2 l = ππππππl = NNNNNNNNNN, (e-3) NNNNNN RR = ρρ e aaaa = ρρ ππππl e aa 2 bb =.9453 2 Ω.9 2 Ω. (e-4) The total power PP of Joule heat generated in the coil is PP = II 2 RR =.9453 6 W =.9 6 W. (e-5) Thus the power density is PP ττ = PP NNNNNNNNNN = PP lππππππ =.7 W/m 3. (e-6)* Note that, by Eqs. (e-3) to (e-5), the expression for power density may also be written as PP ττ = II2 RR ττ = II2 lππππππ ρρ ππππl e aa 2 bb = ρρ e II 2 aaaa = ρρ e JJ 2. (e-7)* This is identical to that obtained in Eq. (e-2). (f) The time rate of temperature increase of the coil is TT = ρρ ejj 2 = ρρ e II 2 ρρ mm cc pp ρρ mm cc pp aaaa. (f-) At TT = 293 K and II =. ka, we have TT = ρρ e II 2 ρρ mm cc pp aaaa = ρρ ejj 2 = 4.975 3 K/s 5. 3 K/s. (f-2)* ρρ mm cc pp (ALTERNATIVE) The heat capacity of the coil is MMcc pp = ρρ mm (lππππππ)cc pp = 3.9 2 J/K 3.9 2 J/K. (f-3) From Eqs. (e-5) and (f-3), the time rate of temperature increase is TT = II2 RR = 4.975 3 K/s 5. 3 K/s. (f-4)* MMcc pp
25 April 2 Page 4 of 6 (Document Released: 4:3, 4/24) Part D. A Pulsed-Field Magnet (g) The magnetic flux φφ BB through each turn is, in the limit l, given by φφ BB = { lim BB()}ππ DD 2 l 2 = μμ 2 II aa ππ DD 2. (g-) The inductance LL of the coil is LL = NNφφ BB = NNμμ 2 II aa ππ DD 2 = lμμ 4aa 2 ππdd2 =.659 4 H. 4 H. (g-2)* The resistance RR of the coil is the same as given in Eq. (e-4). Thus RR = ρρ e ππππππ aaaa = ρρ e ππππl aa 2 bb =.9453 2 Ω.9 2 Ω. (g-3)* (h) According to Kirchhoff s circuit law, the change of electric potential around a closed circuit must be zero and we have LL + RRRR + QQ CC =. (h-) In this question, we are given QQ(tt) = CCVV ee sin(ωωωω + θθ sin θθ ) = ee ( θθ ) CCVV ωω ee (tt+ θθ ) ωω sin ωω(tt + θθ sin θθ ωω ), () II(tt) = = ( cos θθ ) CCVV sin θθ ee sin ωωωω, (2) tan θθ = ωω. (3) Comparing the right sides of Eqs. () and (2), one sees that the current II(tt) = / is obtained from QQ(tt) by changing the latter s time variable tt to (tt θθ /ωω) or, equivalently, changing (tt + θθ /ωω) to tt, and then multiplying its amplitude constant by a factor ee θθ ωω ( ). cos θθ Since II(tt) in Eq. (2) has the same form as QQ(tt) in Eq. (), we may apply the same rule again to obtain its derivative ddii/ as = θθ ee ωω ( ) ( ) CCVV cos θθ cos θθ ee (tt θθ sin θθ ωω ) sin ωω(tt θθ ωω ) = ( ) 2 CCVV ee sin(ωωωω θθ cos θθ sin θθ ) (h-2) Making use of Formula 2 given in Appendix, we may express the left side of Eq. (h-) as a linear combination of cos ωωωω and sin ωωωω so that LL + RRRR + QQ CC = CCVV ee (AA cos θθ sin θθ sin ωωωω + BB sin θθ cos ωωωω) =, (h-3)
25 April 2 Page 5 of 6 (Document Released: 4:3, 4/24) which can be satisfied if and only if AA LL( ) 2 RR( cos θθ cos 2 ) + θθ CC =, (h-4) BB LL( ) 2 + cos θθ CC =, (h-5) Note that Eqs. (h-4) and (h-5) may be obtained more simply by considering Eq. (h-) at the moments when sin ωωωω = and, respectively. Subtracting Eq. (h-5) from Eq. (h-4), we obtain = RR 2LL, If we use the expressions given in Eqs. (g-2) and (g-3), we obtain = RR ππππl ρρ ee 2LL = aa 2 bb 2ρρ ee ππdd 2 lμμ = μμ bbbb = 9.249 s 9. s. 2aa 2 Adding up Eqs. (h-4) and (h-5), we have, by Eq. (h-6) and Eq. (3), LLLL = This may be rewritten as RRRR 2LL cos 2 θθ = 2 cos 2 θθ = 2 ( + tan 2 θθ ) = 2 + ωω 2. (h-6) (h-7)* (h-8) ωω 2 = ωω 2 2 = LLLL ( RR 2LL )2 with ωω = LLLL = 9.7 2 rad/s, (h-9)* and we obtain ωω = ωω 2 2 = 9.6428 2 rad/s 9.6 2 rad/s. (h-)* (i) From Eq. (h-2), the maximum value of II(tt) appears at / = when the time is tt m = θθ ωω. From Eq. (2), the maximum value of II(tt) is then given by II m = II(tt m ) = ( )CCVV cos θθ ee From Eqs. (3), (h-7) and (h-), we have ωω θθ. (i-) (i-2)* tan θθ = ωω =.568, θθ =.4764 rad, tt m = θθ ωω =.53 3 s. (i-3) If II m does not exceed II b found in Problem (d), we must have II m = II(tt m ) II b or ( ) CC VV cos θθ ee ωω θθ II b, (i-4) which implies that the maximum value VV b of VV occurs when the equality holds and is given by VV b = II b ee ωω θθ cos θθ = 2.623 3 V 2. 3 V. (i-5)*
25 April 2 Page 6 of 6 (Document Released: 4:3, 4/24) [j] When II(tt) reaches its maximum at tt = tt m = θθ /ωω, the voltage of the capacitor has dropped from the initial voltage VV = VV b to VV(tt m ) = QQ(tt m) = VV b ee ωω θθ sin(2θθ CC sin θθ ) = 2VV b ee ωω θθ cos θθ. (j-) From tt = to tt = tt m, the energy supplied by the capacitor bank to the circuit, in the form of Joule heat and magnetic energy in the field of the coil, is EE CC = 2 CC VV b 2 [VV(tt m )] 2 = 2 CCVV b 2 4ee 2 ωω θθ cos 2 θθ. (j-2) By the law of conservation of energy, this entire amount of energy is eventually turned into heat in the coil and we have EE = EE CC = 2 CCVV b 2 4ee 2 ωω θθ cos 2 θθ = 2.694 4 J 2. 4 J. (j-3)* If the heat capacity (as computed in Eq. (f-3) remains about the same as that at TT = 293 K, then the temperature increase TT is TT = EE EE = = 53 K. (j-4)* MMcc pp ρρ mm (lππππππ)cc pp With such a temperature increase, the thermal and electrical properties of a metal such as copper do not change substantially.